# Linear depedency

1. May 29, 2009

### evagelos

Prove whethere is linearly independent or not the following:

{$$e^x,e^{2x}$$}

2. May 29, 2009

### jbunniii

If not, then there must exist two scalars, a and b (they can't both be zero), such that

$$a e^x + b e^{2x} = 0$$

for all x. Can you find such scalars?

3. May 29, 2009

### HallsofIvy

Staff Emeritus
Since that is to be true for all x, you can get two simple equations to solve for a and b by choosing two values of x.

4. May 29, 2009

### jeff1evesque

I think it's possible:
$$a e^x + b e^{2x} = 0$$ <==> $$a e^x = - b e^{2x}$$ <==> $$ln (a e^x) = - ln (b e^{2x})$$ <==> ax = -2bx

So if we carefully choose a, we can definitely show your given set is linearly dependent.

5. May 29, 2009

### jbunniii

Except

$$\ln(a e^x) = \ln(a) + x$$
$$\ln(-b e^{2x}) = \ln(-b) + 2x$$

so you need

$$\ln(a) - \ln(-b) = x$$ for all x!

6. May 29, 2009

### jeff1evesque

Bummer, so sorry.

7. May 29, 2009

### evagelos

Put x=0,then a+b=0 ====> a=-b

Hence the above are linearly dependent

CORRECT???

THANKS

8. May 30, 2009

### matt grime

No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.

9. May 30, 2009

### evagelos

Mat grime ,please ,write down for me the definition ,when a set of n vectors are linearly independent

10. May 30, 2009

### matt grime

Apologies, I missed out an 'in'. The functions e^x and e^2x are clearly linearly independent (over R), which is where I misunderstood what you're saying. (If you don't see it then just rearrange ae^x +be^2x=0 to see that you're claim of linear dependence implies that e^x = -a/b for all x.)

Last edited: May 30, 2009
11. May 30, 2009

### HallsofIvy

Staff Emeritus
No, that is incorrect. You have shown that taking a= -b makes $ae^x+ be^{2x}= 0$ for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

Another way to prove it is this: if $ae^x+ be^{2x}= 0$ for all x, then it is a constant and its derivative must be 0 for all x: $ae^x+ 2be^{2x}= 0$. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.

12. May 30, 2009

### evagelos

Is not the definition for linear indepedence the following:

for all a,b and x ,real Nos and $$ae^x + be^{2x}=0\Longrightarrow a=0=b$$

13. May 30, 2009

### HallsofIvy

Staff Emeritus
Yes, that is why you can't "prove $e^x$ and $e^{2x}$ are dependent": they are independent.

14. May 30, 2009

### matt grime

Remember this is equal to the zero *function* in the vector space of real valued functions. The zero function is the one that is zero for all its inputs.

15. May 30, 2009

### evagelos

Then the negation of the above definition should imply for linear dependency

But the negation of the above definition is:

There exist a.b, and ,x such that $$ae^x + be^{2x} = 0$$ and$$a\neq 0$$ and $$b\neq 0$$.

HENCE if we put a=2 ,b=-2 ,x=0 ,we satisfy the linear dependency definition

16. May 31, 2009

### matt grime

That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.

Last edited: May 31, 2009
17. May 31, 2009

### evagelos

Can you put that into a quantifier form??

18. May 31, 2009

### matt grime

In what way is that not in a useful quantifier form (for those who like things like that)? Really, we shouldn't even bother with the quantifier "for all a,b", and putting things in the full on formal abstract quantifier notation just makes things far more opaque than they need to be.

You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.

19. May 31, 2009

### HallsofIvy

Staff Emeritus
The point is that you are working in a vector space of functions. When we say "af(x)+ bg(x)= 0" we mean it is equal to the 0 function- 0 for all x.

20. May 31, 2009

### evagelos

The negation of :

(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

(there exist a,b) [~(ae^x + be^2x = 0 for all x ====> a=b=0)] ======>

(there exist a,b) [ ae^x = be^2x =0 there exist x AND a=/=0 ,b=/=0]

Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.

The negation of $$\forall xPx$$ is : $$\exists x\neg Px$$ .

In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:

ae^x = be^2x =0 and a=/=0,b=/=0

When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

To say that: for all x, is not part of the definition ,then what is part of??